K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
31 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
NT
0
LN
0
DH
3
21 tháng 10 2020
a) \(7x-10=5x-6\)
\(7x-5x=-6+10\)
\(2x=4\)
\(x=2\)
b) \(3x\left(x-2\right)+x-2=0\)
\(\left(x-2\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) \(2x^2+7x-4=0\)
\(2x^2-x+8x-4=0\)
\(x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
21 tháng 10 2020
7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2
3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2 HAY 3x+1=0=>x=-1/3
2x2+7x-4=0.
Câu cuối xem có lộn đề không nha bạn ơi!!!
18 tháng 8 2021
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
18 tháng 8 2021
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
NT
3
27 tháng 7 2018
I don't now
sorry
...................
nha
KT
27 tháng 7 2018
a) \(\left(3x-1\right)^2+2\left(3x-1\right)\left(2x+1\right)+\left(2x+1\right)^2=0\)
\(\Leftrightarrow\)\(\left[\left(3x-1\right)+\left(2x-1\right)\right]^2=0\)
\(\Leftrightarrow\)\(\left(5x-2\right)^2=0\)
\(\Leftrightarrow\)\(5x-2=0\)
\(\Leftrightarrow\)\(x=\frac{2}{5}\)
Vậy...
b) \(\left(7x+2\right)^2+\left(7x-2\right)^2-2\left(7x+2\right)\left(7x-2\right)=0\)
\(\Leftrightarrow\)\(\left[\left(7x+2\right)-\left(7x-2\right)\right]^2=0\)
\(\Leftrightarrow\)\(4^2=0\) vô lí
Vậy pt vô nghiệm
H
2
NN
8 tháng 12 2019
(Ko chép lại đề)
\(a.\Rightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}\)
\(b.\left(3x-2\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=7\end{cases}}\)
\(c.7x^2=28\)
\(x^2=4\)
\(x^2=2^2\)
\(x=\pm2\)
\(d.\left(2x+1\right)\left(1+x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}\)
AD
12 tháng 8 2021
a)\(\left(3x+5\right)\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+5=0\\4-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{4}{3}\end{cases}}}\)
b)\(3x\left(x-7\right)-2\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{2}{3}\end{cases}}}\)
c) \(7x^2-28=0\)
\(\Leftrightarrow7x^2=28\)
\(\Leftrightarrow7x^2=7.4\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
d)\(\left(2x+1\right)+x\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(1+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-1\end{cases}}}\)
#H
VT
2
HT
19 tháng 7 2016
a, \(x^3-7x=0\Leftrightarrow x^2\left(x-7\right)=0\)
\(\left(+\right)x^2=0\Leftrightarrow x=0\)
\(\left(+\right)x-7=0\Leftrightarrow x=7\)
Vậy \(x=0;x=7\)
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)
\(\Leftrightarrow x^3+8-x^3-2x=0\)
\(\Leftrightarrow8-2x=0\)
\(\Leftrightarrow x=4\)
Vậy x=4
\(2x^2-7x-4=0\)
\(2x^2-8x+x-4=0\)
\(2x\left(x-4\right)+\left(x-4\right)=0\)
\(\left(x-4\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\Rightarrow x=4\\2x+1=0\Rightarrow x=-\frac{1}{2}\end{cases}}\)
=.= hok tốt!!